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Collapse of the Brazos River Bridge, Brazos, Texas, during erection of the 973-ft, continuous steel plate girders that support the roadway. The failure was initiated by overstress of the connections between the web and flange during erection. Structures are particularly vulnerable to failure during erection because stiffening elements—for example, floor slabs and bracing—may not be in place. In addition, the structure’s strength may be reduced when certain connections are partially bolted or not fully welded to permit pre- cise alignment of members.

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  • Collapse of the Brazos River Bridge, Brazos, Texas, during erection of the 973-ft, continuous steel plategirders that support the roadway. The failure was initiated by overstress of the connections between theweb and flange during erection. Structures are particularly vulnerable to failure during erection becausestiffening elementsfor example, floor slabs and bracingmay not be in place. In addition, the structuresstrength may be reduced when certain connections are partially bolted or not fully welded to permit pre-cise alignment of members.

  • C H A P T E R

    9Deflections of Beams and Frames

    9.1 Introduction

    When a structure is loaded, its stressed elements deform. In a truss, barsin tension elongate and bars in compression shorten. Beams bend andcables stretch. As these deformations occur, the structure changes shapeand points on the structure displace. Although these deflections are nor-mally small, as part of the total design, the engineer must verify that thesedeflections are within the limits specified by the governing design code toensure that the structure is serviceable. For example, large deflections ofbeams can lead to cracking of nonstructural elements such as plaster ceil-ings, tile walls, or brittle pipes. The lateral displacement of buildings pro-duced by wind forces must be limited to prevent cracking of walls andwindows. Since the magnitude of deflections is also a measure of a mem-bers stiffness, limiting deflections also ensures that excessive vibrationsof building floors and bridge decks are not created by moving loads.

    Deflection computations are also an integral part of a number of ana-lytical procedures for analyzing indeterminate structures, computing buck-ling loads, and determining the natural periods of vibrating members.

    In this chapter we consider several methods of computing deflectionsand slopes at points along the axis of beams and frames. These methodsare based on the differential equation of the elastic curve of a beam. Thisequation relates curvature at a point along the beams longitudinal axisto the bending moment at that point and the properties of the cross sec-tion and the material.

    9.2 Double Integration Method

    The double integration method is a procedure to establish the equationsfor slope and deflection at points along the longitudinal axis (elasticcurve) of a loaded beam. The equations are derived by integrating the

  • differential equation of the elastic curve twice, hence the name doubleintegration. The method assumes that all deformations are produced bymoment. Shear deformations, which are typically less than 1 percent ofthe flexural deformations in beams of normal proportions, are not usu-ally included. But if beams are deep, have thin webs, or are constructedof a material with a low modulus of rigidity (plywood, for example), themagnitude of the shear deformations can be significant and should beinvestigated.

    To understand the principles on which the double integration methodis based, we first review the geometry of curves. Next, we derive the dif-ferential equation of the elastic curvethe equation that relates the cur-vature at a point on the elastic curve to the moment and the flexural stiff-ness of the cross section. In the final step we integrate the differentialequation of the elastic curve twice and then evaluate the constants ofintegration by considering the boundary conditions imposed by the sup-ports. The first integration produces the equation for slope; the secondintegration establishes the equation for deflection. Although the methodis not used extensively in practice since evaluating the constants of inte-gration is time-consuming for many types of beams, we begin our studyof deflections with this method because several other important proce-dures for computing deflections in beams and frames are based on thedifferential equation of the elastic curve.

    Geometry of Shallow Curves

    To establish the geometric relationships required to derive the differen-tial equation of the elastic curve, we will consider the deformations ofthe cantilever beam in Figure 9.1a. The deflected shape is represented inFigure 9.1b by the displaced position of the longitudinal axis (also calledthe elastic curve). As reference axes, we establish an x-y coordinate sys-tem whose origin is located at the fixed end. For clarity, vertical dis-tances in this figure are greatly exaggerated. Slopes, for example, aretypically very smallon the order of a few tenths of a degree. If we wereto show the deflected shape to scale, it would appear as a straight line.

    To establish the geometry of a curved element, we will consider aninfinitesimal element of length ds located a distance x from the fixed end.As shown in Figure 9.1c, we denote the radius of the curved segment by r.At points A and B we draw tangent lines to the curve. The infinitesimalangle between these tangents is denoted by du. Since the tangents to thecurve are perpendicular to the radii at points A and B, it follows that theangle between the radii is also du. The slope of the curve at point Aequals

    dy

    dx tan u

    294 Chapter 9 Deflections of Beams and Frames

    (a)P

    (b)

    y

    x

    x

    A ds B

    ds

    dx

    (c)

    line tangent at B

    line tangent at A

    A

    B

    d

    d

    o

    Figure 9.1

  • Section 9.2 Double Integration Method 295

    If the angles are small (tan u u radians), the slope can be written

    (9.1)

    From the geometry of the triangular segment ABo in Figure 9.1c, wecan write

    (9.2)

    Dividing each side of the equation above by ds and rearranging terms give

    (9.3)

    where du/ds, representing the change in slope per unit length of distancealong the curve, is called the curvature and denoted by the symbol c.Since slopes are small in actual beams, ds dx, and we can express thecurvature in Equation 9.3 as

    (9.4)

    Differentiating both sides of Equation 9.1 with respect to x, we canexpress the curvature du/dx in Equation 9.4 in terms of rectangular coor-dinates as

    (9.5)

    Differential Equation of the Elastic Curve

    To express the curvature of a beam at a particular point in terms of themoment acting at that point and the properties of the cross section, wewill consider the flexural deformations of the small beam segment of lengthdx, shown with darker shading in Figure 9.2a. The two vertical lines rep-resenting the sides of the element are perpendicular to the longitudinal axis

    du

    dx

    d 2y

    dx 2

    c du

    dx

    1r

    c du

    ds

    1r

    r du ds

    dy

    dx u

    dx

    x

    (a)

    d

    N.A.

    (b)

    M

    N.A.

    (c)

    ( f )

    d c

    F

    D

    E

    A

    B

    d

    dl

    dx

    M

    (d)

    (e)

    cFigure 9.2: Flexural deformations of seg-ment dx: (a) unloaded beam; (b) loaded beamand moment curve; (c) cross section of beam;(d ) flexural deformations of the small beamsegment; (e) longitudinal strain; ( f ) flexuralstresses.

  • of the unloaded beam. As load is applied, moment is created, and thebeam bends (see Fig. 9.2b); the element deforms into a trapezoid as thesides of the segment, which remain straight, rotate about a horizontal axis(the neutral axis) passing through the centroid of the section (Fig. 9.2c).

    In Figure 9.2d the deformed element is superimposed on the originalunstressed element of length dx. The left sides are aligned so that thedeformations are shown on the right. As shown in this figure, the longi-tudinal fibers of the segment located above the neutral axis shortenbecause they are stressed in compression. Below the neutral axis the lon-gitudinal fibers, stressed in tension, lengthen. Since the change in lengthof the longitudinal fibers (flexural deformations) is zero at the neutralaxis (N.A.), the strains and stresses at that level equal zero. The variationof longitudinal strain with depth is shown in Figure 9.2e. Since the strainis equal to the longitudinal deformations divided by the original lengthdx, it also varies linearly with distance from the neutral axis.

    Considering triangle DFE in Figure 9.2d, we can express the changein length of the top fiber dl in terms of du and the distance c from theneutral axis to the top fiber as

    (9.6)

    By definition, the strain P at the top surface can be expressed as

    (9.7)

    Using Equation 9.6 to eliminate dl in Equation 9.7 gives

    (9.8)

    Using Equation 9.5 to express the curvature du/dx in rectangular coordi-nates, we can write Equation 9.8 as

    (9.9)

    If behavior is elastic, the flexural stress, s, can be related to the strainP at the top fiber by Hookes law, which states that

    where E the modulus of elasticity

    Solving for P gives

    (9.10)

    Using Equation 9.10 to eliminate P in Equation 9.9 produces

    (9.11)d 2y

    dx 2s

    Ec

    P s

    E

    s EP

    d 2y

    dx 2

    Pc

    P du

    dx c

    P dl

    dx

    dl du c

    296 Chapter 9 Deflections of Beams and Frames

  • Section 9.2 Double Integration Method 297

    For elastic behavior the relationship between the flexural stress at the topfiber and the moment acting on the cross section is given by

    (5.1)

    Substituting the value of s given by Equation 5.1 into Equation 9.11 pro-duces the basic differential equation of the elastic curve

    (9.12)

    In Examples 9.1 and 9.2 we use Equation 9.12 to establish the equa-tions for both the slope and the deflection of the elastic curve of a beam.This operation is carried out by expressing the bending moment in termsof the applied load and distance x along the beams axis, substituting theequation for moment in Equation 9.12, and integrating twice. Themethod is simplest to